Diazonium salts — preparation, Sandmeyer, Gattermann, coupling reactions (azo dyes): Worked examples

Definition Two symbols we will lean on all page

$Δ$ (capital Greek delta) over a reaction arrow means "heat" — you warm the flask, usually with a proper hot bath or flame. So $Δ$ reads literally as "on heating, this becomes that." We reserve it for steps that genuinely need a heat source; merely letting a solution drift above $5^{\circ} C$ is not $Δ$ — we will say "warm gently" for that, so you never confuse mild warming with strong heating.
EAS = electrophilic aromatic substitution. "Electrophile" = an electron-loving species; "aromatic" = a benzene-type ring; "substitution" = one atom on the ring (usually an $H$ ) is swapped out for the incoming group. So EAS means an electron-hungry attacker replaces a hydrogen on a benzene ring. See Electrophilic Aromatic Substitution for the full mechanism. This is the engine of the coupling reaction (cell C8).

The scenario matrix

Every diazonium problem falls into one of these case classes. Think of the columns as "what do you want on the ring?" and the rows as "what family of trick does that need?" The figure below colour-codes the classes so you can see the three families at a glance before reading the table.

Figure — Diazonium salts — preparation, Sandmeyer, Gattermann, coupling reactions (azo dyes)

Figure — the nine case classes sorted into three coloured families: violet = build the salt (C1), orange = fire the rocket and plug a new group in the empty seat (C2–C7), magenta = keep both nitrogens and make colour (C8), with navy for the traps (C9). Every worked example below is tagged with its cell so you can map it back here.

Legend for the coloured dots in the table below: 🟣 = violet family (build the salt), 🟠 = orange family (fire the rocket, plug a new group), 🔴 = magenta family (keep both N's, make colour), 🔵 = navy family (traps / degenerate inputs). These match the four colours of the figure exactly.

#	Case class	Trigger words in the question	Rocket ( $N_{2}$ ) fired?	Governing idea
C1	🟣 violet — Make the salt (diazotisation)	"primary aromatic amine + $N a N O_{2} / H C l$ , $0 - 5^{\circ} C$ "	not yet — we build it	resonance keeps it alive when cold
C2	🟠 orange — Cl / Br / CN on ring — Sandmeyer	"with $C u C l$ / $C u B r$ / $C u C N$ "	yes (radical)	Cu(I) makes the aryl radical
C3	🟠 orange — Cl / Br on ring — Gattermann	"Cu powder + HX"	yes (radical)	Cu made in situ, cheaper
C4	🟠 orange — I on ring	"just $K I$ , warm"	yes	$I^{-}$ works with no copper
C5	🟠 orange — F on ring — Balz–Schiemann	" $H B F_{4}$ , then heat"	yes	fluoroborate salt isolated first
C6	🟠 orange — $O H$ on ring (phenol)	"warm in water / dil. acid, $> 5^{\circ} C$ "	yes	water is the nucleophile
C7	🟠 orange — $H$ on ring (deamination)	" $H_{3} P O_{2}$ or $E tO H$ "	yes	replace group by hydrogen
C8	🔴 magenta — Azo coupling (dye)	"with phenol / aniline, mild pH"	NO — both N's kept	EAS, weak electrophile
C9	🔵 navy — Degenerate / trap inputs	aliphatic amine, secondary amine, wrong temperature, wrong pH	—	why the reaction fails

The 9 worked examples below hit every cell at least once. The cell each example covers is printed in its title.

Worked examples

Worked example Example 1 — Cell C1 (build the salt) + C9 (degenerate: why not aliphatic)

Statement. You have two amines: aniline ( $C_{6} H_{5} - N H_{2}$ , an amine on a benzene ring) and ethylamine ( $C H_{3} C H_{2} - N H_{2}$ , an amine on a plain carbon chain). You treat each with $N a N O_{2} + H C l$ at $0 - 5^{\circ} C$ . Which one gives an isolable diazonium salt, and why?

Forecast: guess now — same reagents, both primary amines, so… same outcome? (Careful.)

Identify the electrophile. $N a N O_{2} + 2 H C l \to N O^{+} + C l^{-} + N a C l + H_{2} O$ . The active species is the nitrosonium ion $N O^{+}$ ("nitroso-cation," a nitrogen–oxygen unit carrying $+ 1$ ). Why this step? Nothing happens until we know what actually attacks the nitrogen — it is $N O^{+}$ , not $H N O_{2}$ directly.
Both amines react with $N O^{+}$ to give a diazonium cation of the form $(carbon skeleton) - N +_{2}$ — for aniline this is $C_{6} H_{5} - N +_{2}$ , for ethylamine $C H_{3} C H_{2} - N +_{2}$ . Why this step? The first stage is identical; the difference only shows up in what happens next. (We deliberately avoid a generic "R" symbol here so nothing undefined sneaks in — recall $Ar$ means specifically an aromatic ring, which the ethyl chain is not.)
Aniline case: the benzene ring's $π$ -system shares (delocalises) the positive charge onto the ring by resonance, so $C_{6} H_{5} N +_{2} C l^{-}$ survives at $0 - 5^{\circ} C$ . Why this step? Spreading out a charge lowers energy — the salt is stable enough to isolate. Look at the figure below: the $+$ charge is drawn hopping between the nitrogen and several ring positions — a charge shared over many atoms is a charge no single atom has to bear, and that is what "resonance-stabilised" means.
Ethylamine case: a plain carbon chain has no $π$ -system to share the charge. The $- N +_{2}$ immediately fires: $C H_{3} C H_{2} - N +_{2} \to C H_{3} C H_{2}^{+} + N_{2} ↑$ , giving a messy mix (ethanol, ethene). Why this step? Without stabilisation the rocket fires instantly — no salt to collect. In the figure, the alkyl case has the $+$ stuck on one nitrogen with nowhere to spread — high energy, so it blows apart.

Figure — left: aryl diazonium, the magenta $+$ charge is delocalised across the ring (several resonance positions), so the ion is calm and survives. Right: alkyl diazonium, the $+$ is trapped on one atom (no ring), so it immediately ejects $N_{2}$ .

Verify: Nitrogen balance in the aniline product: $C_{6} H_{5} N H_{2}$ has 1 N; product $C_{6} H_{5} N +_{2} C l^{-}$ has 2 N. The extra N came from $N a N O_{2}$ — exactly one N added. ✓ Charge balance: $N +_{2}$ is $+ 1$ , $C l^{-}$ is $- 1$ , net neutral salt. ✓

Worked example Example 2 — Cell C2 (Sandmeyer, halogen) + C3 contrast (Gattermann)

Statement. Starting from aniline, give two different reagent routes to bromobenzene ( $C_{6} H_{5} B r$ ), and state which gives higher yield.

Forecast: both routes put Br on the ring — will they use the same copper reagent?

Diazotise first (cell C1): $C_{6} H_{5} N H_{2} N a N O_{2} / H C l, 0 - 5^{\circ} C C_{6} H_{5} N +_{2} C l^{-}$ . Why this step? The unreactive $- N H_{2}$ can't be swapped for Br directly; convert it to the excellent leaving group $- N +_{2}$ first.
Route A (Sandmeyer): add $C u B r$ (a soluble copper(I) salt). $C_{6} H_{5} N +_{2} C l^{-} C u B r C_{6} H_{5} B r + N_{2} ↑ + (Cl^{-} stays in solution)$ . Why this step? Cu(I) donates one electron to the diazonium, creating an aryl radical $C_{6} H_{5}^{∙}$ after $N_{2}$ leaves; the Cu–Br then hands Br to that radical. (The original $C l^{-}$ counter-ion is a spectator — it does not enter the product.)
Route B (Gattermann): add Cu metal powder + HBr. Same product: $C_{6} H_{5} N +_{2} C l^{-} C u / H B r C_{6} H_{5} B r + N_{2} ↑$ . Why this step? Cheap copper metal generates Cu(I) in situ, doing the same radical job without pre-made $C u B r$ .
Compare. Sandmeyer (pure $C u B r$ ) gives higher yield; Gattermann is cheaper but lower yield. Why this step? Exams love asking which you'd pick — the trade-off is yield vs. cost/convenience.

Verify: Write the full balanced equation keeping every spectator: $C_{6} H_{5} N +_{2} C l^{-} + C u B r \to C_{6} H_{5} B r + N_{2} + C u C l$ . Check each element — C: 6=6, H: 5=5, N: 2=2, Br: 1=1, Cu: 1=1, Cl: 1=1 (the chloride migrates to the copper, it is not lost). ✓ Exactly one $N_{2}$ released. ✓

Worked example Example 3 — Cell C4 (iodide, no copper)

Statement. From benzenediazonium chloride, make iodobenzene ( $C_{6} H_{5} I$ ). Someone hands you $C u I$ "to match Sandmeyer." Do you need it?

Forecast: halogen going on → surely we need copper like Br and Cl?

Recall the special case. Iodide is a large, easily-oxidised (electron-rich) ion. $C_{6} H_{5} N +_{2} C l^{-} + K I warm gently C_{6} H_{5} I + N_{2} ↑ + K C l$ . Why this step? $I^{-}$ can itself supply the electron the copper normally supplies — it kicks off the radical chain without any copper catalyst.
Reject the $C u I$ . No copper is needed; adding it is a wasted reagent (and a classic trap). Why this step? The whole point of this cell is that $I$ breaks the "halogens need Cu" pattern.

Verify: Balance: $C_{6} H_{5} N +_{2} C l^{-} + K I \to C_{6} H_{5} I + N_{2} + K C l$ . K: 1=1, I: 1=1, Cl: 1=1, N: 2=2, C6H5 both sides. ✓

Worked example Example 4 — Cell C5 (Balz–Schiemann, fluoride)

Statement. Convert aniline to fluorobenzene ( $C_{6} H_{5} F$ ). Why can't you just add $K F$ (analogous to $K I$ )?

Forecast: if $I$ works with $K I$ , does $F$ work with $K F$ ?

Diazotise, but with $H B F_{4}$ (fluoroboric acid). This gives the diazonium tetrafluoroborate $C_{6} H_{5} N +_{2} B F_{4}^{-}$ , a solid you can isolate. Why this step? The $B F_{4}^{-}$ counter-ion carries the fluorine you'll install and makes a stable, dry salt.
Heat the dry salt (thermal decomposition). Here $Δ$ genuinely means a heat source (recall the $Δ$ = heat definition at the top — this really is strong heating of a dry solid, not mild warming): $C_{6} H_{5} N +_{2} B F_{4}^{-} Δ C_{6} H_{5} F + N_{2} ↑ + B F_{3}$ . Why this step? On heating, $N_{2}$ leaves and $F$ from the fluoroborate plugs the seat — this is the Balz–Schiemann reaction.
Why not just $K F$ ? $F^{-}$ is small, tightly held, and a poor electron donor — it won't run the radical chain like $I^{-}$ does, and in water it's a weak nucleophile. Why this step? This is the exam's "feels-symmetric-but-isn't" trap ( $K I$ works, $K F$ does not).

Verify: Balance the decomposition: $C_{6} H_{5} N +_{2} B F_{4} \to C_{6} H_{5} F + N_{2} + B F_{3}$ . F: LHS 4 = RHS (1 in $C_{6} H_{5} F$ ) $+$ (3 in $B F_{3}$ ) $= 4$ . ✓ B: 1=1, N: 2=2. ✓

Worked example Example 5 — Cell C6 (phenol) + the temperature trap from C9

Statement. p-toluidine (4-methylaniline, $4 - C H_{3} - C_{6} H_{4} - N H_{2}$ ) is diazotised, then the aqueous solution is warmed gently above $5^{\circ} C$ . What forms — and how is this the same chemistry that we spent Example 1 trying to prevent?

Forecast: in prep we feared warming; here we choose it. Contradiction?

Diazotise cold (cell C1): $4 - C H_{3} - C_{6} H_{4} - N H_{2} N a N O_{2} / H C l, 0 - 5^{\circ} C 4 - C H_{3} - C_{6} H_{4} - N +_{2} C l^{-}$ . Why this step? Build the salt while cold so it survives.
Now warm it gently in water (just above $5^{\circ} C$ — no strong heat source needed): $4 - C H_{3} - C_{6} H_{4} - N +_{2} C l^{-} + H_{2} O warm gently 4 - C H_{3} - C_{6} H_{4} - O H + N_{2} ↑ + H C l$ . The product is p-cresol (4-methylphenol). Why this step? Warm water attacks the ring as $- N +_{2}$ leaves as $N_{2}$ ; the seat is taken by $- O H$ . Note we write "warm gently," not $Δ$ — this decomposition happens with only mild warming, which is exactly why it is a hazard during preparation.
Reconcile with Example 1. This hydrolysis is exactly the decomposition we feared during preparation. The lesson: cold = keep the salt; warm gently = deliberately destroy it into phenol. Why this step? Same reaction, opposite intention — temperature is the switch.

Verify: Balance: $C H_{3} C_{6} H_{4} N +_{2} C l^{-} + H_{2} O \to C H_{3} C_{6} H_{4} O H + N_{2} + H C l$ . N: 2=2, Cl: 1=1, O: 1 (water) → 1 (in $O H$ ). H count: LHS ring $C H_{3} C_{6} H_{4}$ = C7H7, plus $H_{2} O$ = 2H → 9 H on left of the organic+water; RHS $C H_{3} C_{6} H_{4} O H$ = C7H8O (8 H) + HCl (1 H) = 9 H. ✓

Worked example Example 6 — Cell C7 (deamination: replace

$- N H_{2}$ by $- H$ ) Statement. You must make 1,3,5-tribromobenzene from benzene. Direct bromination of benzene stops well before three Br's go on symmetrically. The clean trick uses aniline. Show the last step: how do you remove the nitrogen after it has done its job?

Forecast: we install a group then throw it away — what reagent turns $- N +_{2}$ into just $- H$ ?

Use $- N H_{2}$ as a strong director. Aniline's $- N H_{2}$ is a powerful ortho/para activator (see Activating and Directing Groups); brominating aniline gives 2,4,6-tribromoaniline cleanly. Why this step? We need a group that forces bromine into the 2,4,6 pattern first.
Diazotise (cell C1): $2, 4, 6 - B r_{3} C_{6} H_{2} - N H_{2} N a N O_{2} / H C l, 0 - 5^{\circ} C 2, 4, 6 - B r_{3} C_{6} H_{2} - N +_{2} C l^{-}$ . Why this step? Turn the now-unwanted $- N H_{2}$ into the leaving group.
Deaminate with $H_{3} P O_{2}$ (hypophosphorous acid) in water. Writing the complete equation with every by-product: $2, 4, 6 - B r_{3} C_{6} H_{2} - N +_{2} C l^{-} + H_{3} P O_{2} + H_{2} O \to 1, 3, 5 - B r_{3} C_{6} H_{3} + N_{2} ↑ + H_{3} P O_{3} + H C l$ . Why this step? $H_{3} P O_{2}$ delivers an H to the aryl radical, replacing $- N +_{2}$ by $- H$ ; the phosphorus is oxidised from hypophosphorous acid ( $H_{3} P O_{2}$ ) to phosphorous acid ( $H_{3} P O_{3}$ ) — that is where the "used-up" reducing agent goes, and it must appear for the atoms to balance.

Verify: Balance the full equation $B r_{3} C_{6} H_{2} N_{2} C l + H_{3} P O_{2} + H_{2} O \to B r_{3} C_{6} H_{3} + N_{2} + H_{3} P O_{3} + H C l$ . Br: 3=3. C: 6=6. N: 2=2. Cl: 1=1. P: 1=1. O: LHS 2 (in $H_{3} P O_{2}$ ) + 1 (water) = 3 = RHS 3 (in $H_{3} P O_{3}$ ). ✓ H: LHS = 2 (ring $C_{6} H_{2}$ ) + 3 ( $H_{3} P O_{2}$ ) + 2 ( $H_{2} O$ ) = 7; RHS = 3 (ring $C_{6} H_{3}$ ) + 3 ( $H_{3} P O_{3}$ ) + 1 (HCl) = 7. ✓

Verify (product symmetry): Product $1, 3, 5 - B r_{3} C_{6} H_{3}$ has formula $C_{6} H_{3} B r_{3}$ : 3 ring H's + 3 Br's on 6 carbons, symmetric. Starting aniline positions 2,4,6 (Br) map to 1,3,5 (Br) after removing the N from position 1. ✓

Worked example Example 7 — Cell C8 (azo coupling, the dye) — geometric/electronic

Statement. Benzenediazonium chloride is added to phenol in mild $N a O H$ . Draw/name the product, say where on the phenol ring it couples and why, and explain why it is coloured. Confirm no $N_{2}$ is lost.

Forecast: every reaction so far dumped $N_{2}$ . Will this one too?

Set the pH (cell C8 rule). Mild $N a O H$ converts phenol to phenoxide $C_{6} H_{5} O^{-}$ , whose $- O^{-}$ is an even stronger ortho/para activator than $- O H$ . Why this step? A more electron-rich ring is needed because the diazonium is only a weak electrophile — and the coupling proceeds by EAS (electrophilic aromatic substitution, defined at the top of this page; see also Electrophilic Aromatic Substitution). Too much base, though, would destroy the diazonium — hence mild.
Couple at the para position. The bulky $Ar- N +_{2}$ attacks the ring; ortho is crowded, so it lands para to the $- O^{-}$ . Why this step? $- O^{-}$ directs o/p; para is the less hindered of the two for a large electrophile.
Keep both nitrogens. The new bond forms from the ring carbon to the terminal N of $- N +_{2}$ ; both N's stay, giving the azo bridge $- N = N -$ . Product: p-hydroxyazobenzene, $C_{6} H_{5} - N = N - C_{6} H_{4} - O H$ (an orange dye). Why this step? This is EAS — the electrophile is retained whole; no $N_{2}$ is expelled, unlike C2–C7.
Why coloured? The chain $C_{6} H_{5} - N = N - C_{6} H_{4} - O H$ is one long conjugated $π$ -system (see Conjugation and Colour — Chromophores). Long conjugation shrinks the gap between the highest occupied and lowest empty $π$ energy levels until that gap matches visible light; the molecule absorbs part of the visible spectrum and we see the complementary colour. Why this step? This links the structure directly to "it's a dye."

Read the figure below as you follow the steps: the violet ring on the left comes from the diazonium; the magenta $- N = N -$ in the middle is the retained bridge (a stamp on the right reminds you no $N_{2}$ is released); the orange ring on the right is the phenol, with the new bond landing para to its $- O H$ ; the double-headed magenta arrow across the top marks the single long conjugated span that makes the molecule absorb visible light and appear orange.

Figure — azo coupling of benzenediazonium with phenol: the two rings are joined at the para position through a retained $- N = N -$ bridge, forming one continuous conjugated pi-system (the orange chromophore); crucially, both nitrogens stay, so no $N_{2}$ leaves.

Verify: Nitrogen bookkeeping: reactant $C_{6} H_{5} N +_{2}^{+}$ has 2 N; product azo bridge $- N = N -$ has 2 N; 0 N lost. ✓ This is the one cell where $N_{2}$ is retained. ✓

Worked example Example 8 — Cell C9 (exam twist: wrong pH kills the coupling)

Statement. A student tries to couple benzenediazonium chloride with aniline but works in strongly acidic solution and gets no dye. Diagnose the failure and give the correct condition.

Forecast: more acid = more reactive, right? Guess whether the dye forms.

What strong acid does to aniline. Excess $H^{+}$ protonates the amine: $C_{6} H_{5} N H_{2} + H^{+} \to C_{6} H_{5} N + H_{3}$ . The resulting $- N + H_{3}$ is a deactivating, electron-withdrawing group. Why this step? Coupling needs an electron-rich ring; the protonated ring is electron-poor and refuses the weak electrophile.
What strong base would do instead. Swing the other way and add too much $O H^{-}$ : the diazonium is destroyed — $C_{6} H_{5} N +_{2}^{+} + 2 O H^{-} \to C_{6} H_{5} - N = N - O^{-} + H_{2} O$ , an unreactive diazotate that no longer acts as an electrophile (and warm base pushes on toward phenol). Why this step? It shows the coupling window is squeezed from both sides — acid kills the ring, base kills the diazonium — so we know the answer must lie in between.
Correct condition: mildly acidic to neutral pH for amine coupling — enough free (unprotonated) aniline to stay reactive, but not enough base to kill the diazonium. Product then: 4-aminoazobenzene $C_{6} H_{5} - N = N - C_{6} H_{4} - N H_{2}$ (yellow). Why this step? Pinpoints the narrow "Goldilocks" pH window the exam is testing.

Verify: Logic check — free aniline ring is activated ( $- N H_{2}$ is $o / p$ -activator), protonated ring is deactivated ( $- N + H_{3}$ withdraws). The active reactant is the neutral amine, so neither extreme pH works. Nitrogen bookkeeping on the successful product: reactant diazonium 2 N $+$ aniline 1 N $= 3$ N; product $C_{6} H_{5} - N = N - C_{6} H_{4} - N H_{2}$ has bridge (2 N) $+$ amine (1 N) $= 3$ N, 0 lost — consistent with a coupling (cell C8). ✓ Consistent with the parent note's "narrow pH window." ✓

Worked example Example 9 — Cell C9 (degenerate input: a secondary amine gives no diazonium at all)

Statement. A student diazotises N-methylaniline ( $C_{6} H_{5} - N H - C H_{3}$ , a secondary aromatic amine — the nitrogen carries the ring and a methyl group and only one H) with $N a N O_{2} / H C l$ at $0 - 5^{\circ} C$ , expecting a diazonium salt like plain aniline gives. What actually forms, and why can no diazonium salt appear?

Forecast: it is still an aromatic amine on a benzene ring — surely it diazotises like aniline?

Count the N–H bonds. Diazotisation needs a primary amine: two N–H bonds, because building $- N + \equiv N$ from $- N H_{2}$ requires stripping both hydrogens off the nitrogen (recall the parent mechanism $Ar-NH_{2} \to Ar-NH-N=O \to Ar- N +_{2}$ ). Why this step? If the nitrogen has a carbon group where a second H should be, that pathway simply cannot complete.
See what N-methylaniline has instead. Its nitrogen has only one N–H (the other position is taken by $- C H_{3}$ ). So $N O^{+}$ still attacks the nitrogen, but it stops at an N-nitroso compound: $C_{6} H_{5} - N (C H_{3}) - H + N O^{+} \to C_{6} H_{5} - N (C H_{3}) - N = O + H^{+}$ . Why this step? With no second H to remove, the reaction cannot dehydrate on to the triple-bonded $- N +_{2}$ ; it freezes at N-nitroso-N-methylaniline (a yellow oil), not a diazonium salt.
State the rule. Only primary aromatic amines ( $- N H_{2}$ ) give diazonium salts; secondary aromatic amines give N-nitrosamines; the reaction is not a "failed diazotisation" but a different reaction entirely. Why this step? This is the exact C9 trap — same-looking reagents, wrong amine class, completely different product.

Verify: N-balance for the actual product: reactant amine has 1 N; $N O^{+}$ brings 1 N; N-nitroso product $C_{6} H_{5} N (C H_{3}) N = O$ contains 2 N — one from the amine, one from $N O^{+}$ , none lost. ✓ H-check on step 2: the single N–H is removed as the $H^{+}$ shown, and no second N–H exists, so the diazonium ( $- N +_{2}$ , which would need both H's gone) can never form. ✓

Recall Quick self-test on the matrix

Which cell fires the $N_{2}$ rocket AND needs copper? ::: C2 (Sandmeyer, for Cl/Br/CN) and C3 (Gattermann, Cu powder) — both radical, both lose $N_{2}$ . Which single cell does NOT lose $N_{2}$ ? ::: C8, azo coupling — both nitrogens stay as the $- N = N -$ bridge. Halogen that needs no copper at all? ::: Iodine (cell C4), just $K I$ and warm gently. Reagent to replace $- N +_{2}$ by plain $- H$ , and its phosphorus by-product? ::: $H_{3} P O_{2}$ (hypophosphorous acid); it is oxidised to $H_{3} P O_{3}$ (phosphorous acid) — cell C7, deamination. Why does warming during preparation ruin the salt but warming later is useful? ::: Warming (even gently) hydrolyses $- N +_{2}$ to $- O H$ (phenol, cell C6) — bad in prep, but exactly what we want for phenol synthesis. What do a primary vs secondary aromatic amine give on diazotisation? ::: Primary → diazonium salt; secondary → N-nitrosamine (cell C9 trap). What does $Δ$ over a reaction arrow mean, and when do we NOT use it? ::: "Heat" from a real heat source; we do NOT use it for merely warming a solution above 5 °C (that is "warm gently"). What does EAS stand for? ::: Electrophilic aromatic substitution — an electron-hungry species replaces an H on a benzene ring.